Answer
See below
Work Step by Step
a) Let $v=(v_1,v_2,v_3)\\
u=(u_1,u_2,u_3)\in R^3$
Obtain: $Ker (T)=\{v \in R^3: T(v)=0\}\\
=\{v \in R^3: T(u,v)=(0,0)\\
=\{(v_1,v_2,v_3)\in R^3:(u_1v_1+u_2v_2+u_3v_3=0)$
Since $u$ is a fixed nonzero vector in $R^3$, we can see that at least one of $u_1,u_2,u_3$ will not equal to $0$
Assume that $u_1$ is not equal to $0$, we have:
$u_1v_1+u_2v_2+u_3v_3=0\\
\rightarrow v_1=-\frac{u_2v_2+u_3v_3}{u_1}$
Thus, $Ker (T)=\{(-\frac{u_2v_2+u_3v_3}{u_1},v_2,v_3):v_2,v_3 \in R)\}\\
=\{v_2(-\frac{u_2}{u_1},1,0)+v_3(-\frac{u_3}{u_1},0,1):v_2,v_3 \in R\\
=span \{(-\frac{u_2}{u_1},1,0),(-\frac{u_3}{u_1},0,1)\}$ $\dim [Ker (T)]=2\\
=\{ v\in R^3:=0\}\\$
$=\{v\in R^3: v$ is orthogonal to $u\}$
Geometrically, $Ker(T)$ is a set of all vectors orthogonal to the given nonzero vector $u$.
b) Then, $Rng(T)= \{T(v):v\in R^3\}\\
=\{: v \in R^3\\
=\{ a\in R:=a,v \in R^3$
